Evaluate $\lim\limits_{x\rightarrow \infty}(1+\frac{1}{\sqrt{x}})^{\sqrt{x}}$.

Can I get some help? I am thinking that the limit does not exist. If you approach it from the left and then from the right, I think that the limits do not equal each other. I also suspect that we are dealing with Euler's limit, i.e.

The limits $\lim\limits_{x\rightarrow \infty +} (1+\frac{1}{x})^x$ and $\lim\limits_{x\rightarrow \infty -} (1+\frac{1}{x})^x$ exist and both equal $e$

Otherwise, I have little intuition to go from.

  • $\begingroup$ I edited that out, assuming it was a mistake. Apologies OP if that is incorrect. $\endgroup$ – James Apr 21 '15 at 14:26
  • $\begingroup$ $\sqrt{x}$ does not exist when $x<0$ and $+\infty$ can be approached only from the left. $\endgroup$ – Jack D'Aurizio Apr 21 '15 at 14:27
  • $\begingroup$ What does $\infty+$ and $\infty-$ mean? Do you mean $+\infty, -\infty$? $\endgroup$ – GFauxPas Apr 21 '15 at 14:29
  • 1
    $\begingroup$ Note that in real analysis $\infty$ is not a real point. We merely use the notation $lim_{x \to \infty} f(x) = L$ to mean $\forall \epsilon > 0 \exists M \in \mathbb{R}: \forall x \in \mathbb{R}\ x > M \implies | f(x) - L | < \epsilon$. This differs from the usual limit in that, instead of getting closer and closer to some value $L$ as we approach a point $a$, we now get closer and closer to some value $L$ as we make $x$ bigger and bigger. Consequently left- and right-limits do not exist for $\infty$. $\endgroup$ – Eric Spreen Apr 21 '15 at 14:38
  • $\begingroup$ In many books (imo, in most of them), writing the limit when $\;x\to\infty\;$ means $\;x\;$ approaches plus infinity , and there's only one way to approach this: from the left. For minus infinity we write $\;x\to -\infty\;$ $\endgroup$ – Timbuc Apr 21 '15 at 14:39

When $\;x\to\infty\;$ we can assume $\;x>0\;$ when doing the limit, so now simply make a substitution:

$$x\leftrightarrow y^2\;\implies\;\;x\to\infty\iff y\to\infty$$

and your limit becomes


For negative $\;x$'s $\;\sqrt x\;$ isn't defined and thus also the limit of your expression isn't when $\;x\to -\infty\;$


Note that we can write

\begin{equation*} \lim_{x\to\infty} e^{(\sqrt{x}\ln(1+\frac{1}{\sqrt{x}}))} = e^{(\lim_{x\to\infty} \sqrt{x}\ln(1+\frac{1}{\sqrt{x}}))} =e^{\left(\lim_{x\to\infty}\frac{\ln(1+\frac{1}{\sqrt{x}})}{\frac{1}{\sqrt{x}}}\right)}. \end{equation*}

Applying L'Hopital's rule rule gives

\begin{equation*} e^{\left(\lim_{x\to\infty}\frac{1}{1+\frac{1}{\sqrt{x}}}\right)}=e. \end{equation*}


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